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STRUCTURE OF Ne-20 AND Ne-19
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Despite the enormous success of the Bohr model (1913) and the quantum mechanics of Schrodinger (1926) based on the well-established laws of electromagnetism in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, after the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) neither was able to discover the nuclear structure, because the discovery of the assumed uncharged neutron (1932) led to the abandonment of electromagnetic laws in favour of wrong theories, which could not lead to the correct nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks, discovered by Gell-Mann and Zweig, in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for the discovery of nuclear force and structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Using the following diagrams of the structures of F-19 and Ne-20 and also Fig.8d and Fig. 6c of my published paper one sees that the total spins are the fundamental characteristics which lead to the nuclear structures of Ne-20 with S = 0 and of F-19 with S = +1/2. Both F19 and Ne19 have the same number of extra nucleons which make rectangles outside the stable parallelepiped of O-16 (core). Under this condition we also discovered why the unstable Ne-19 with S = +1/2 turns to the stable structure of F19 with a similar spin S = +1/2 . ' ' Nuclear structure of the stable Ne-20 with S=0 ' In Fig.6c of my published paper you see that the structure of Ne-20 is based on the very stable structure of O16 with S = 0 which plays the role of a core. It is the well-known parallelepiped with the fundamental four horizontal planes. The nucleons of the second (HP2) and third plane (HP3) form a parallelepiped of a great stability because it is characterized by points with four pn bonds per nucleon. This is the central parallelepiped of the structure of O-16 which is responsible for the stability of O-16 with B(O16) = -127.58 MeV. In the following second diagram of Ne20 you see that the extra p9 , n9 , p10 and n10 of the second and the third horizontal plane (HP2) respectively form two extra stable rectangle of S = 0 which give the total S=0 of the structure of Ne20 with a binding energy B(Ne) = - 160.62 MeV. Thus B(Ne) – B(O16) = (-160.62) – (127.58) = -33.04 MeV It means that the axial and radial pn bonds of the two extra rectangles overcome the pp and nn repulsions. Here the strong axial and radial pn bonds of extra rectangles are due to the fact that at the second and third horizontal square of the central parallelepiped of the O16 (core ) the extra nucleons contribute to the increase of the binding energies of the pn bonds because they form five pn bonds per nucleon. For example at point n3 of the second horizontal square (n3p3n4p4) one observes the five bonds like n3p1, n3p3, n3p4, n3p5 , and n3p9 . Here p9 does not turn into a neutron because both bonds like p9n3 and p9n9 have a binding energy of short range which overcomes the repulsive energy Q of the long-ranged pp repulsions. Note that the repulsive energy of nn systems is negligible because the repulsive forces of the nn systems are of short range. Therefore the stable structure of two extra rectangles contribute to the stability of Ne20 with S=0. ' ''' '''Stable F19 with S = +1/2 Stable Ne20 with S = 0 p8(-1/2).n8(-1/2) p8(-1/2).n8(-1/2) n7(-1/2).p7(-1/2) n7(-1/2).p7(-1/2) HP4 n6(+1/2).p6(+1/2)….n10(+1/2) n6(+1/2).p6(+1/2)....n10(+1/2) p5(+1/2).n5(+1/2) n9(+1/2)...p5(+1/2).n5(+1/2) HP3 p4(-1/2).n4(-1/2)…..p9(-1/2) p4(-1/2).n4(-1/2)….p10(-1/2) n3 (-1/2).p3 (- 1/2). p9(-1/2)…n3(-1/2).p3(-1/2) HP2 n2(+1/2).p2(+1/2)….n9(+1/2) n2(+1/2).p2(+1/2) p1(+1/2). n1(+1/2) p1(+1/2).n1(+1/2) HP1 ' ' Why the unstable Ne-19 with S = +1/2 turns to the stable F-19 with S = +1/2 ''' '''Looking carefully at the first diagram and the Fig.8d of my published paper one concludes that the three extra nucleons n9(+1/2) , p9(-1/2) and n10(+1/2) of the F19 give three radial and two axial bonds with S = +1/2 explaining the total spin S= +1/2 of the structure of F19 ( See my STRUCTURE OF F19, F18 AND O18 ). Such bonds contribute to the increase of the short-ranged binding energies of the pn systems, which overcome the repulsive energy Q of the pp repulsions of long range . For example the stable bonds of p10n4 and p10n10 of the Ne20, are due to the increased binding energies as a result of the five pn bonds per nucleon at point n4. In the same way in F19 the extra p9 does not turn to a neutron because the p9 makes the fifth bond at the same point n4 which overcome the repulsive energy Q of the pp repulsions of long range. It is of interest to note that the unstable structure of Ne-19 with S = +1/2 which is similar to the spin of F19 has the same number of thee extra nucleons like p9 (+1/2), n9(-1/2) and p10(+1/2) by replacing the n9, p9 and n10 of the stable F19 respectively. In the same way here they form three radial and two axial bonds. (On purpose such bonds are not shown here because they form the same extra rectangles as those of F19). As in the case of F19 one concludes that they are coupled with n1, p3 and n5 of the structure of O16 respectively. So there exist the three radial bonds like p9n1, n9p3 and p10n5 and the two axial bonds like p9n9 and n9p10 . However the n1p9 bond in the unstable N19 is not stable because at point n1 we do not observe five bonds per nucleon but four ones . Thus the pp repulsions of long range with a repulsive energy Q are able to overcome the n1p9 bond. Under this condition the p9 turns into a neutron n which represents the neutron n9 of the stable F19 as p9 + Q = n9 + positron + neutrino Note that in the absence of long- ranged nn repulsions in the stable F19, the new neutron like the n9 makes the stable n9p2 and n9p9 bonds, while at the same first horizontal square the p9 should turn into the new neutron n9 because in the unstable Ne19 the repulsive energy Q of the pp repulsions of long range could overcome the four bonds per nucleon. Category:Fundamental physics concepts